# How I can break a long tikz box in an automatic way?

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{shapes,snakes}
\usepackage{amsmath,amssymb}
\begin{document}

% Define box and box title style
\tikzstyle{mybox} = [draw=red, fill=blue!20, very thick,
rectangle, rounded corners, inner sep=10pt, inner ysep=20pt]
\tikzstyle{fancytitle} =[fill=red, text=white]

\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{1\textwidth}
To calculate the horizontal position the kinematic differential
equations are needed:
\begin{align}
\dot{n} &= u\cos\psi -v\sin\psi \\
\dot{e} &= u\sin\psi + v\cos\psi
\end{align}
For small angles the following approximation can be used:
\begin{align}
\dot{n} &= u -v\delta_\psi \\
\dot{e} &= u\delta_\psi + v
\end{align}‎
‎\begin{itemize}
‎\item‎‎
1‎
‎\item‎‎
2‎‎
‎\item‎‎
1‎
‎\item‎‎
2‎‎
‎\item‎‎
1‎
‎\item‎‎
2‎‎
‎\item‎‎
1‎
‎\item‎‎
2‎‎
‎\item‎‎
1‎
‎\item‎‎
2‎
\end{itemize} ‎‎
To calculate the horizontal position the kinematic differential
equations are needed:
\begin{align}
\dot{n} &= u\cos\psi -v\sin\psi \\
\dot{e} &= u\sin\psi + v\cos\psi
\end{align}
For small angles the following approximation can be used:
\begin{align}
\dot{n} &= u -v\delta_\psi \\
\dot{e} &= u\delta_\psi + v
\end{align}‎‎
To calculate the horizontal position the kinematic differential
equations are needed:
\begin{align}
\dot{n} &= u\cos\psi -v\sin\psi \\
\dot{e} &= u\sin\psi + v\cos\psi
\end{align}
For small angles the following approximation can be used:
\begin{align}
\dot{n} &= u -v\delta_\psi \\
\dot{e} &= u\delta_\psi + v
\end{align}‎‎
To calculate the horizontal position the kinematic differential
equations are needed:
\begin{align}
\dot{n} &= u\cos\psi -v\sin\psi \\
\dot{e} &= u\sin\psi + v\cos\psi
\end{align}
For small angles the following approximation can be used:
\begin{align}
\dot{n} &= u -v\delta_\psi \\
\dot{e} &= u\delta_\psi + v
\end{align}‎‎
\end{minipage}‎

};
\node[fancytitle, right=10pt] at (box.north west) {A fancy title};
\node[fancytitle, rounded corners] at (box.east) {$\clubsuit$};
\end{tikzpicture}%

\end{document}


I want to write a long box (more than one page) by tikz. it is possible that break box automatically?

I'd suggest you to use the powerful tcolorbox package to produce your box allowing page breaks; here's the translation of your box using tcolorbox's settings (adjust the parameters according to your needs):

\documentclass{article}
\usepackage[many]{tcolorbox}
\usepackage{amsmath,amssymb}

\newtcolorbox{myfancybox}{
breakable,
enhanced jigsaw,
colback=blue!20,
colframe=red,
top=12pt,
overlay unbroken={\node[fill=red,text=white] at (frame.east) {$\clubsuit$};},
attach boxed title to top left={xshift=10pt,yshift=-8pt},
boxed title style={size=small,colback=red,colframe=red},
title=#1
}

\begin{document}

\begin{myfancybox}{The title}
To calculate the horizontal position the kinematic differential
equations are needed:
\begin{align}
\dot{n} &= u\cos\psi -v\sin\psi \\
\dot{e} &= u\sin\psi + v\cos\psi
\end{align}
\end{myfancybox}

\begin{myfancybox}{The title}
To calculate the horizontal position the kinematic differential
equations are needed:
\begin{align}
\dot{n} &= u\cos\psi -v\sin\psi \\
\dot{e} &= u\sin\psi + v\cos\psi
\end{align}
For small angles the following approximation can be used:
\begin{align}
\dot{n} &= u -v\delta_\psi \\
\dot{e} &= u\delta_\psi + v
\end{align}‎
‎\begin{itemize}
\item Test
\end{itemize} ‎‎
To calculate the horizontal position the kinematic differential
equations are needed:
\begin{align}
\dot{n} &= u\cos\psi -v\sin\psi \\
\dot{e} &= u\sin\psi + v\cos\psi
\end{align}
For small angles the following approximation can be used:
\begin{align}
\dot{n} &= u -v\delta_\psi \\
\dot{e} &= u\delta_\psi + v
\end{align}‎‎
To calculate the horizontal position the kinematic differential
equations are needed:
\begin{align}
\dot{n} &= u\cos\psi -v\sin\psi \\
\dot{e} &= u\sin\psi + v\cos\psi
\end{align}
For small angles the following approximation can be used:
\begin{align}
\dot{n} &= u -v\delta_\psi \\
\dot{e} &= u\delta_\psi + v
\end{align}‎‎
To calculate the horizontal position the kinematic differential
equations are needed:
\begin{align}
\dot{n} &= u\cos\psi -v\sin\psi \\
\dot{e} &= u\sin\psi + v\cos\psi
\end{align}
For small angles the following approximation can be used:
\begin{align}
\dot{n} &= u -v\delta_\psi \\
\dot{e} &= u\delta_\psi + v
\end{align}‎‎
\end{myfancybox}%

\end{document}


The result: 